Three vectors of equal magnitude are shown in figure. Find angle θ formed by \(\vec{a} + \vec{b} + \vec{c}\) with x axis, tanθ = ?
(a) \(\frac{\sqrt{6}+1}{\sqrt{2}-1}\)
(b) \(\frac{\sqrt{6}-1}{\sqrt{2}+1}\)
(c) \(\frac{1}{\sqrt 2}\)
(d) \(\frac{\sqrt{2}-1}{\sqrt{6}+1}\)
Correct answer: (a) \(\frac{\sqrt{6}+1}{\sqrt{2}-1}\)
Explanation:
Magnitude of each vector be
\(\vec{a}\) = \(\frac{x}{2}\hat{i}\) + \(\frac{x\sqrt{3}}{2}\hat{j}\)
\(\vec{b}\) = \(\frac{x}{\sqrt 2}\hat{i}\) + \(\frac{x}{\sqrt 2}\hat{j}\)
\(\vec{a}\) = -\(\frac{x}{2}\hat{i}\) + \(\frac{x\sqrt{3}}{2}\hat{j}\)
\(\vec{a} + \vec{b} - \vec{c}\) = (\(x -\frac{x}{\sqrt 2}\hat{i}\)) + (\(x \sqrt{3} + \frac{x}{\sqrt 2}\hat{j}\))
tanθ = \(\frac{\sqrt 3 + \frac{1}{\sqrt 2}}{1 - \frac{1}{\sqrt 2}}\)
= \(\frac{\sqrt{6}+1}{\sqrt{2}-1}\)